Q: Find the value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$.

$$ \begin{aligned} \cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) & =\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right] \\ & =\cos ^{-1}\left[\cos \frac{\pi}{6}\right] \\ & =\frac{\pi}{6} \end{aligned} $$

Q: Find the value of $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$.

$$ \begin{aligned} \tan ^{-1}\left(\tan \frac{7 \pi}{6}\right) & =\tan ^{-1}\left[\tan \left(2 \pi-\frac{5 \pi}{6}\right)\right] \\ & =\tan ^{-1}\left[-\tan \left(\frac{5 \pi}{6}\right)\right] \\ & =\tan ^{-1}\left[\tan \left(\pi-\frac{5 \pi}{6}\right)\right] \\ & =\tan ^{-1}\left[\tan \frac{\pi}{6}\right] \\ & =\frac{\pi}{6} \end{aligned} $$

Q: Prove that $\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$

$$ \begin{aligned} \text { LHS } & =\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8} \\ & =\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \cdot \frac{1}{7}}\right)+\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \cdot \frac{1}{8}}\right) \\ & =\tan ^{-1}\left(\frac{12}{34}\right)+\tan ^{-1}\left(\frac{11}{23}\right) \\ & =\tan ^{-1}\left(\frac{6}{17}\right)+\tan ^{-1}\left(\frac{11}{23}\right) \\ & =\tan ^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \cdot \frac{11}{23}}\right) \\ & =\tan ^{-1}\left(\frac{325}{325}\right) \\ & =\tan ^{-1}(1) \\ & =\frac{\pi}{4} \\ & =\text { RHS } \end{aligned} $$

Q: Prove that $\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]$.

Let $x=\tan ^{2} \theta$ Then, $$ \begin{aligned} \sqrt{x} & =\tan \theta \\ \theta & =\tan ^{-1} \sqrt{x} \end{aligned} $$ Therefore, $$ \begin{aligned} \left(\frac{1-x}{1+x}\right) & =\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta} \\ & =\cos 2 \theta \end{aligned} $$ Thus, $$ \begin{aligned} R H S & =\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) \\ & =\frac{1}{2} \cos ^{-1}(\cos 2 \theta) \\ & =\frac{1}{2} \times 2 \theta \\ & =\theta \\ & =\tan ^{-1} \sqrt{x} \\ & =L H S \end{aligned} $$

Question 1

Find the value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$.

Accepted Answer

$$

\begin{aligned}

\cos ^{-1}\left(\cos \frac{13 \pi}{6}ight) & =\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}ight)ight] \

& =\cos ^{-1}\left[\cos \frac{\pi}{6}ight] \

& =\frac{\pi}{6}

\end{aligned}

$$

Question 2

Find the value of $	an ^{-1}\left(	an \frac{7 \pi}{6}ight)$.

Accepted Answer

$$

\begin{aligned}

an ^{-1}\left(	an \frac{7 \pi}{6}ight) & =	an ^{-1}\left[	an \left(2 \pi-\frac{5 \pi}{6}ight)ight] \

& =	an ^{-1}\left[-	an \left(\frac{5 \pi}{6}ight)ight] \

& =	an ^{-1}\left[	an \left(\pi-\frac{5 \pi}{6}ight)ight] \

& =	an ^{-1}\left[	an \frac{\pi}{6}ight] \

& =\frac{\pi}{6}

\end{aligned}

$$

Question 3

Prove that $2 \sin ^{-1} \frac{3}{5}=	an ^{-1} \frac{24}{7}$.

Accepted Answer

Let $\sin ^{-1} \frac{3}{5}=x \Rightarrow \sin x=\frac{3}{5}$

Then,

$$

\cos x=\sqrt{1-\left(\frac{3}{5}ight)^{2}}=\frac{4}{5}

$$

Therefore,

$$

\begin{align*}

an x & =\frac{3}{4} \

x & =	an ^{-1} \frac{3}{4} \

\sin ^{-1} \frac{3}{5} & =	an ^{-1} \frac{3}{4} 	ag{1}

\end{align*}

$$

Thus,

$$

\begin{aligned}

L H S & =2 \sin ^{-1} \frac{3}{5} \

& =2 	an ^{-1} \frac{3}{4} \

& =	an ^{-1}\left(\frac{2 	imes \frac{3}{4}}{1-\left(\frac{3}{4}ight)^{2}}ight) \quad[	ext { from }(1)] \

& =	an ^{-1}\left(\frac{24}{7}ight) \

& =R H S

\end{aligned}

$$

Question 4

Prove that $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=	an ^{-1} \frac{77}{36}$.

Accepted Answer

Let $\sin ^{-1} \frac{8}{17}=x \Rightarrow \sin x=\frac{8}{17}$

Then,

$$

\cos x=\sqrt{1-\left(\frac{8}{17}ight)^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}

$$

Therefore,

$$

\begin{align*}

an x & =\frac{8}{15} \

x & =	an ^{-1} \frac{8}{15} \

\sin ^{-1} \frac{8}{17} & =	an ^{-1} \frac{8}{15} 	ag{1}

\end{align*}

$$

Now, let $\sin ^{-1} \frac{3}{5}=y \Rightarrow \sin y=\frac{3}{5}$

Then,

$$

\cos y=\sqrt{1-\left(\frac{3}{5}ight)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}

$$

Therefore,

$$

\begin{align*}

an y & =\frac{3}{4} \

y & =	an ^{-1} \frac{3}{4} \

\sin ^{-1} \frac{3}{5} & =	an ^{-1} \frac{3}{4} 	ag{2}

\end{align*}

$$

Thus, by using (1) and (2)

$$

\begin{aligned}

ext { LHS } & =\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} \

& =	an ^{-1} \frac{8}{15}+	an ^{-1} \frac{3}{4} \

& =	an ^{-1}\left[\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \cdot \frac{3}{4}}ight] \

& =	an ^{-1}\left[\frac{\frac{32+45}{60}}{\frac{60-24}{60}}ight] \

& =	an ^{-1} \frac{77}{36} \

& =	ext { RHS }

\end{aligned}

$$

Question 5

Prove that $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$.

Accepted Answer

Let $\cos ^{-1} \frac{4}{5}=x \Rightarrow \cos x=\frac{4}{5}$

Then,

$$

\sin x=\sqrt{1-\left(\frac{4}{5}ight)^{2}}=\frac{3}{5}

$$

Therefore,

$$

\begin{align*}

an x & =\frac{3}{4} \

x & =	an ^{-1} \frac{3}{4} \

\cos ^{-1} \frac{4}{5} & =	an ^{-1} \frac{3}{4} 	ag{1}

\end{align*}

$$

Now, let $\cos ^{-1} \frac{12}{13}=y \Rightarrow \cos y=\frac{12}{13}$

Then,

$$

\sin y=\frac{5}{13}

$$

Therefore,

$$

\begin{align*}

an y & =\frac{5}{12} \

y & =	an ^{-1} \frac{5}{12} \

\cos ^{-1} \frac{12}{13} & =	an ^{-1} \frac{5}{12} 	ag{2}

\end{align*}

$$

Thus, by using (1) and (2)

$$

\begin{align*}

\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13} & =	an ^{-1} \frac{3}{4}+	an ^{-1} \frac{5}{12} \

& =	an ^{-1}\left[\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}ight] \

& =	an ^{-1}\left[\frac{56}{33}ight] 	ag{3}

\end{align*}

$$

Now, let $\cos ^{-1} \frac{33}{65}=z \Rightarrow \cos z=\frac{33}{65}$

Then,

$$

\sin z=\sqrt{1-\left(\frac{33}{65}ight)^{2}}=\frac{56}{65}

$$

Therefore,

$$

\begin{align*}

an z & =\frac{33}{56} \

z & =	an ^{-1} \frac{56}{33} \

\cos ^{-1} \frac{33}{65} & =	an ^{-1} \frac{56}{33} 	ag{4}

\end{align*}

$$

Thus, by using (3) and (4)

$$

\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}

$$

Hence proved.

Question 6

Prove that $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$.

Accepted Answer

Let $\cos ^{-1} \frac{12}{13}=y \Rightarrow \cos y=\frac{12}{13}$

Then,

$$

\sin y=\sqrt{1-\left(\frac{12}{13}ight)^{2}}=\frac{5}{13}

$$

Therefore,

$$

\begin{align*}

an y & =\frac{5}{12} \

y & =	an ^{-1} \frac{5}{12} \

\cos ^{-1} \frac{12}{13} & =	an ^{-1} \frac{5}{12} 	ag{1}

\end{align*}

$$

Now, let $\sin ^{-1} \frac{3}{5}=x \Rightarrow \sin x=\frac{3}{5}$

Then,

$$

\cos x=\sqrt{1-\left(\frac{3}{5}ight)^{2}}=\frac{4}{5}

$$

Therefore,

$$

\begin{align*}

an x & =\frac{3}{4} \

x & =	an ^{-1} \frac{3}{4} \

\sin ^{-1} \frac{3}{5} & =	an ^{-1} \frac{3}{4} 	ag{2}

\end{align*}

$$

Now, let $\sin ^{-1} \frac{56}{65}=z \Rightarrow \sin z=\frac{56}{65}$

Then,

$$

\cos z=\sqrt{1-\left(\frac{56}{65}ight)^{2}}=\frac{33}{65}

$$

Therefore,

$$

\begin{align*}

& 	an z=\frac{56}{33} \

& z=	an ^{-1} \frac{56}{33} \

& \sin ^{-1} \frac{56}{65}=	an ^{-1} \frac{56}{33} 	ag{3}

\end{align*}

$$

Thus, by using (1) and (2)

$$

\begin{aligned}

L H S & =\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5} \

& =	an ^{-1} \frac{5}{12}+	an ^{-1} \frac{3}{4} \

& =	an ^{-1}\left[\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}}ight] \

& =	an ^{-1}\left[\frac{\frac{20+36}{\frac{48}{48-15}}}{48}ight] \

& =	an ^{-1}\left(\frac{56}{33}ight) \

& =\sin ^{-1} \frac{56}{65} \quad \quad[	ext { Using }(3)] \

& =R H S \quad

\end{aligned}

$$

Question 7

Prove that $	an ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$.

Accepted Answer

Let $\sin ^{-1} \frac{5}{13}=x \Rightarrow \sin x=\frac{5}{13}$

Then,

$$

\cos x=\sqrt{1-\left(\frac{5}{13}ight)^{2}}=\frac{12}{13}

$$

Therefore,

$$

\begin{align*}

an x & =\frac{5}{12} \

x & =	an ^{-1} \frac{5}{12} \

\sin ^{-1} \frac{5}{13} & =	an ^{-1} \frac{5}{12} 	ag{1}

\end{align*}

$$

Now, let $\cos ^{-1} \frac{3}{5}=y \Rightarrow \cos y=\frac{3}{5}$

Then,

$$

\sin y=\sqrt{1-\left(\frac{3}{5}ight)^{2}}=\frac{4}{5}

$$

Therefore,

$$

\begin{align*}

an y & =\frac{4}{3} \

y & =	an ^{-1} \frac{4}{3} \

\cos ^{-1} \frac{3}{5} & =	an ^{-1} \frac{4}{3} 	ag{2}

\end{align*}

$$

Thus, by using (1) and (2)

$$

\begin{aligned}

ext { RHS } & =\sin ^{-1} \frac{5}{12}+\cos ^{-1} \frac{3}{5} \

& =	an ^{-1} \frac{5}{12}+	an ^{-1} \frac{4}{3} \

& =	an ^{-1}\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \cdot \frac{4}{3}}ight) \

& =	an ^{-1}\left(\frac{63}{16}ight) \

& =	ext { LHS }

\end{aligned}

$$

Question 8

Prove that $	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{7}+	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{8}=\frac{\pi}{4}$

Accepted Answer

$$

\begin{aligned}

ext { LHS } & =	an ^{-1} \frac{1}{5}+	an ^{-1} \frac{1}{7}+	an ^{-1} \frac{1}{3}+	an ^{-1} \frac{1}{8} \

& =	an ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \cdot \frac{1}{7}}ight)+	an ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \cdot \frac{1}{8}}ight) \

& =	an ^{-1}\left(\frac{12}{34}ight)+	an ^{-1}\left(\frac{11}{23}ight) \

& =	an ^{-1}\left(\frac{6}{17}ight)+	an ^{-1}\left(\frac{11}{23}ight) \

& =	an ^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \cdot \frac{11}{23}}ight) \

& =	an ^{-1}\left(\frac{325}{325}ight) \

& =	an ^{-1}(1) \

& =\frac{\pi}{4} \

& =	ext { RHS }

\end{aligned}

$$

Question 9

Prove that $	an ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}ight), x \in[0,1]$.

Accepted Answer

Let $x=	an ^{2} 	heta$

Then,

$$

\begin{aligned}

\sqrt{x} & =	an 	heta \

heta & =	an ^{-1} \sqrt{x}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\left(\frac{1-x}{1+x}ight) & =\frac{1-	an ^{2} 	heta}{1+	an ^{2} 	heta} \

& =\cos 2 	heta

\end{aligned}

$$

Thus,

$$

\begin{aligned}

R H S & =\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}ight) \

& =\frac{1}{2} \cos ^{-1}(\cos 2 	heta) \

& =\frac{1}{2} 	imes 2 	heta \

& =	heta \

& =	an ^{-1} \sqrt{x} \

& =L H S

\end{aligned}

$$

Question 10

Prove that $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$.

Accepted Answer

$$

\begin{aligned}

\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}ight) & =\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}} \quad \quad \quad 	ext { (by rationalizing) } \

& =\frac{(1+\sin x)+(1-\sin x)+2 \sqrt{(1+\sin x)(1-\sin x)}}{1+\sin x-1+\sin x} \

& =\frac{2\left(1+\sqrt{1-\sin ^{2} x}ight)}{2 \sin x}=\frac{1+\cos x}{\sin x} \

& =\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \

& =\cot \frac{x}{2}

\end{aligned}

$$

Thus,

$$

\begin{aligned}

L H S & =\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}ight) \

& =\cot ^{-1}\left(\cot \frac{x}{2}ight) \

& =\frac{x}{2} \

& =	ext { RHS }

\end{aligned}

$$

Question 11

Prove that $	an ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}ight)=\frac{\pi}{2}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1$

Accepted Answer

Let $x=\cos 2 	heta \Rightarrow 	heta=\frac{1}{2} \cos ^{-1} x$

Thus,

$$

\begin{aligned}

L H S & =	an ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}ight) \

& =	an ^{-1}\left(\frac{\sqrt{1+\cos 2 	heta}-\sqrt{1-\cos 2 	heta}}{\sqrt{1+\cos 2 	heta}+\sqrt{1-\cos 2 	heta}}ight) \

& =	an ^{-1}\left(\frac{\sqrt{2 \cos ^{2} 	heta}-\sqrt{2 \sin ^{2} 	heta}}{\sqrt{2 \cos ^{2} 	heta}+\sqrt{2 \sin ^{2} 	heta}}ight) \

& =	an ^{-1}\left(\frac{\sqrt{2} \cos 	heta-\sqrt{2} \sin 	heta}{\sqrt{2} \cos 	heta+\sqrt{2} \sin 	heta}ight) \

& =	an ^{-1}\left(\frac{\cos 	heta-\sin 	heta}{\cos 	heta+\sin 	heta}ight) \

& =	an ^{-1}\left(\frac{1-	an 	heta}{1+	an 	heta}ight) \

& =	an ^{-1} 1-	an (	an 	heta) \

& =\frac{\pi}{4}-	heta \

& =\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x \

& =	ext { RHS }

\end{aligned}

$$

Question 12

Prove that $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

Accepted Answer

$$

\begin{align*}

L H S & =\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3} \

& =\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}ight) \

& =\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}ight) 	ag{1}

\end{align*}

$$

Now, let $\cos ^{-1} \frac{1}{3}=x \Rightarrow \cos x=\frac{1}{3}$

Therefore,

$$

\begin{align*}

\sin x & =\sqrt{1-\left(\frac{1}{3}ight)^{2}} \

& =\frac{2 \sqrt{2}}{3} \

x & =\sin ^{-1} \frac{2 \sqrt{2}}{3} \

\cos ^{-1} \frac{1}{3} & =\sin ^{-1} \frac{2 \sqrt{2}}{3} 	ag{2}

\end{align*}

$$

Thus, by using (1) and (2)

$$

\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}

$$

Hence proved.

Question 13

Solve $2 	an ^{-1}(\cos x)=	an ^{-1}(2 \operatorname{cosec} x)$.

Accepted Answer

It is given that $2 	an ^{-1}(\cos x)=	an ^{-1}(2 \operatorname{cosec} x)$

Since, $2 	an ^{-1}(x)=	an ^{-1} \frac{2 x}{1-x^{2}}$

Hence,

$$

\begin{aligned}

& \Rightarrow 	an ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}ight)=	an ^{-1}(2 \operatorname{cosec} x) \

& \Rightarrow\left(\frac{2 \cos x}{1-\cos ^{2} x}ight)=(2 \operatorname{cosec} x) \

& \Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x} \

& \Rightarrow \cos x=\sin x \

& \Rightarrow 	an x=1 \

& \Rightarrow 	an x=	an \frac{\pi}{4}

\end{aligned}

$$

Therefore,

$$

x=n \pi+\frac{\pi}{4}, 	ext { where } n \in Z

$$

Question 14

Solve $	an ^{-1} \frac{1-x}{1+x}=\frac{1}{2} 	an ^{-1} x,(x>0)$

Accepted Answer

Since $	an ^{-1} x-	an ^{-1} y=	an ^{-1} \frac{x-y}{1+x y}$

Hence,

$$

\begin{aligned}

& \Rightarrow 	an ^{-1} \frac{1-x}{1+x}=\frac{1}{2} 	an ^{-1} x \

& \Rightarrow 	an ^{-1} 1-	an ^{-1} x=\frac{1}{2} 	an ^{-1} x \

& \Rightarrow \frac{\pi}{4}=\frac{3}{2} 	an ^{-1} x \

& \Rightarrow 	an ^{-1} x=\frac{\pi}{6} \

& \Rightarrow x=	an \frac{\pi}{6} \

& \Rightarrow x=\frac{1}{\sqrt{3}}

\end{aligned}

$$

Question 15

Solve $\sin \left(	an ^{-1} xight),|x|<1$ is equal to

(A) $\frac{x}{\sqrt{1-x^{2}}}$

(B) $\frac{1}{\sqrt{1-x^{2}}}$

(C) $\frac{1}{\sqrt{1+x^{2}}}$

(D) $\frac{x}{\sqrt{1+x^{2}}}$

Accepted Answer

Let $	an y=x$

Therefore,

$$

\sin y=\frac{x}{\sqrt{1+x^{2}}}

$$

Now, let $	an ^{-1} x=y$

Therefore,

$$

y=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}ight)

$$

Hence,

$$

an ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}ight)

$$

Thus,

$$

\begin{aligned}

\sin \left(	an ^{-1} xight) & =\sin \left(\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}ight)ight) \

& =\frac{x}{\sqrt{1+x^{2}}}

\end{aligned}

$$

Thus, the correct option is D.

Question 16

Solve: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$, then $x$ is equal to

(A) $0, \frac{1}{2}$

(B) $1, \frac{1}{2}$

(C) 0

(D) $\frac{1}{2}$

Accepted Answer

It is given that $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$

$$

\begin{align*}

& \Rightarrow \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \

& \Rightarrow-2 \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1}(1-x) \

& \Rightarrow-2 \sin ^{-1} x=\cos ^{-1}(1-x) 	ag{1}

\end{align*}

$$

Let $\sin ^{-1} x=y \Rightarrow \sin y=x$

Hence,

$$

\begin{aligned}

\cos y & =\sqrt{1-x^{2}} \

y & =\cos ^{-1}\left(\sqrt{1-x^{2}}ight) \

\sin ^{-1} x & =\cos ^{-1} \sqrt{1-x^{2}}

\end{aligned}

$$

From equation (1), we have

$$

-2 \cos ^{-1} \sqrt{1-x^{2}}=\cos ^{-1}(1-x)

$$

Put $x=\sin y$

$$

\begin{aligned}

& \Rightarrow-2 \cos ^{-1} \sqrt{1-\sin ^{2} y}=\cos ^{-1}(1-\sin y) \

& \Rightarrow-2 \cos ^{-1}(\cos y)=\cos ^{-1}(1-\sin y) \

& \Rightarrow-2 y=\cos ^{-1}(1-\sin y) \

& \Rightarrow 1-\sin y=\cos (-2 y) \

& \Rightarrow 1-\sin y=\cos 2 y \

& \Rightarrow 1-\sin y=1-2 \sin ^{2} y \

& \Rightarrow 2 \sin ^{2} y-\sin y=0 \

& \Rightarrow \sin y(2 \sin y-1)=0 \

& \Rightarrow \sin y=0, \frac{1}{2}

\end{aligned}

$$

Therefore,

$$

x=0, \frac{1}{2}

$$

When $x=\frac{1}{2}$, it does not satisfy the equation.

Hence, $x=0$ is the only solution

Thus, the correct option is C.

Question 17

Solve $	an ^{-1}\left(\frac{x}{y}ight)-	an ^{-1} \frac{x-y}{x+y}$ is equal to

(A) $\frac{\pi}{2}$

(B) $\frac{\pi}{3}$

(C) $\frac{\pi}{4}$

(D) $\frac{-3 \pi}{4}$

Accepted Answer

$$

\begin{aligned}

an ^{-1}\left(\frac{x}{y}ight)-	an ^{-1} \frac{x-y}{x+y} & =	an ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}ight)\left(\frac{x-y}{x+y}ight)}ight] \

& =	an ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}ight] \

& =	an ^{-1}\left(\frac{x^{2}+x y-x y+y^{2}}{x y+y^{2}+x^{2}-x y}ight) \

& =	an ^{-1}(1) \

& =	an ^{-1}\left(	an \frac{\pi}{4}ight) \

& =\frac{\pi}{4}

\end{aligned}

$$

Thus, the correct option is C.